Better way to write this in Ruby?

I'm new to Ruby. After a ton of refactoring I came down to this. Is there a better way to wr开发者_如何学Pythonite this?

 51   def tri_num?(n)
 52     i = 1
 53     while i < n
 54       return i if i * (i + 1) / 2 == n
 55       i += 1
 56     end 
 57     raise InvalidTree
 58   end

What about solving it directly?

def tri_num? n

  i = (0.5*(-1.0 + Math.sqrt(1.0 + 8.0*n))).to_i

  if i*(i+1)/2 == n
    return i
  else
    raise InvalidTree
  end

end

Though I don't know if tri_num? is a good name. Usually a function ending with a ? should return true or false.

Yes.

def tri_num?(n)
  1.upto(n-1) do |i|
    return i if i * (i + 1) / 2 == n
  end
  raise InvalidTree
end

I thought the same as dantswain, basically invert the equation:

=> i * (i + 1) / 2 = n
=> i * (i + 1) = 2*n
=> i^2 + i = 2*n
=> i^2 + i -2*n = 0

And the solutions for the above are:

i = (-1 +- sqrt(1+8n))/2

Here I don't consider the - solution as it will give negative for any value of n bigger than 0, in the end the code is:

def tri_num?(n)
    i = (-1 + Math.sqrt(1 + 8*n))/2.0
    return i.to_i if i == i.to_i
    raise InvalidTree
end

def tri_num?(n)
  (1...n).each do |i|
    return i if i * (i + 1) / 2 == n
  end
  rails InvalidTree # not defined..                                             
end